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### Errors In Approximate Integration And Tabular Form

Find the approximation value of Integral in which function is given in tabular form method. Practice error bound for midpoint & trapezoidal rule & simpson rule error.

# Error Bound For Midpoint Rule:

• Let $$E_M$$ be the error while we calculate the value of integral using Midpoint Rule.
• Midpoint Rule says that,

In this method of approximation, we write

• $$\int\limits_a^bf(x)dx\simeq M_{n}=\Delta x \Bigg[f(\overline x_1)+f(\overline x_2) + .........f(\overline x_n)\Bigg]$$

where, $$\Delta\,x=\dfrac{b-a} {n}$$and  $$\overline x_i=\dfrac{x_{i–1}+x_i}{2}\rightarrow$$ midpoint of $$[x_{i–1},x_i]$$

Interval [a,b] is divided into n equal parts each measuring $$\dfrac{b-a}{n}$$=$$\Delta u$$ and then height  $$f(\overline x)$$ is taken of the midpoint of each interval.  • Let  $$|f"(x)|\leq k$$ for $$a\leq x\leq b$$ then,

$$|E_M| \leq \dfrac{k(b-a)^3}{24\,n^2}$$  •   The R.H.S value in this inequality is the maximum error that can creep in
• In general, the actual error is substantiality less than maximum error.

• Let  $$|f"(x)|\leq k$$ for $$a\leq x\leq b$$ then,

$$|E_M| \leq \dfrac{k(b-a)^3}{24\,n^2}$$  •   The R.H.S value in this inequality is the maximum error that can creep in
• In general, the actual error is substantiality less than maximum error.

#### Estimate the error while calculating M8 for the integral $$I = \int\limits ^4_1 \dfrac{1}{\sqrt x} dx$$ .

A $$(E_M) \leq.2652$$

B $$(E_M) \leq 0.01318$$

C $$(E_M) \leq 0.00526$$

D $$(E_M) \geq 1.326$$

×

For the Midpoint Rule

$$|E_M| = Error \leq \dfrac{k(b-a)^3}{24\,n^2}$$    where,  $$|f"(x)|\leq k$$ for $$a\leq x\leq b$$

For the integral $$I = \int\limits^b_a f(x) dx$$ and $$\Delta x = \dfrac{b-a}{n}$$

In this case,

$$f(x) = \dfrac{1}{\sqrt x} = x^{-{1}/{2}}$$

$$= f'(x) = -\dfrac{1}{2} \, x ^{-{3}/{2}}$$

$$\Rightarrow f'' (x) = -\dfrac{1}{2} × -\dfrac{3}{2}× x^{-{5}/{2}} = \dfrac{3}{4}\,x^{-{5}/{2}}$$

a= 1, b=4

The maximum value of $$f"(x)$$  in [1,4] is $$\dfrac{3}{4}$$

$$\Rightarrow k= \dfrac{3}{4}$$

$$\therefore \, |E_M| \leq \,\dfrac{3}{4} \,\dfrac{(4-1)^3}{24× 64} \\= \dfrac{3}{4}× \dfrac{27}{24× 64}\\ = 0.01318$$

$$\therefore\, |E_M| \leq \, 0.01318$$

### Estimate the error while calculating M8 for the integral $$I = \int\limits ^4_1 \dfrac{1}{\sqrt x} dx$$ .

A

$$(E_M) \leq.2652$$

.

B

$$(E_M) \leq 0.01318$$

C

$$(E_M) \leq 0.00526$$

D

$$(E_M) \geq 1.326$$

Option B is Correct

# Error Bounds For Trapezoidal Rule

• Let ET be the error while we calculate the value of integral using Trapezoidal  Rule.
• Trapezoidal  Rule says that,

In this rule of approximation ,

$$\int\limits_a^b\,f(x)\,dx \simeq T_n = \dfrac{\Delta x}{2} \Bigg[f(x_0) + 2f(x_1) + 2f(x_2) + .......+ 2f(x_{n–1}) + f(x_n)\Bigg]$$

where, $$\Delta x = \dfrac {b-a}{n},x_i=a+i\Delta x$$  Each shaded portion is a trapezoidal and we add all the trapezium areas.

For n=5,

Shaded area = $$=\dfrac{1}{2}\bigg(f(x_0) + f(x_1)\bigg) \Delta x + \dfrac {1}{2} \bigg(f(x_1) + f(x_2)\bigg) \Delta x + ..........+ \dfrac {1}{2} \bigg(f(x_n) + f(b)\bigg)\Delta x$$

$$=\dfrac{\Delta x}{2}\Bigg[\underbrace {f(x_0)}_{f(a)} + 2f(x_1) + ..............+ 2f(x_4) + \underbrace {f(x_5)}_{f(b)}\Bigg]$$

• Let $$|f''(x)| \leq k\,\, \text{for}\,\,a\leq x\leq b\,\,\text{then}$$

$$|E_T| \leq \dfrac{k(b-a)^3}{12\,n^2}$$

• The R.H.S. value in the inequality  is the maximum value of error  that can creep in .

• In general the actual error is substantially  less than the maximum error.

#### Estimate the error ,while calculating T8 for the integral $$I = \int\limits ^4_1 \dfrac{1}{x^2} dx$$.

A $$|E_T| \leq 0.0156$$

B $$|E_T| \leq 0.2109$$

C $$|E_T| \leq 0.258$$

D $$|E_T| \leq 5.16$$

×

For the Trapezoidal Rule,

$$|E_T| = Error \leq \dfrac{k(b-a)^3}{12\,n^2}$$

where, $$|f''(x)| \leq k\,\, \text{for}\,\,a\leq x\leq b$$

For the integral , $$I=\int\limits^b_a f(x) dx$$  and $$\Delta x = \dfrac{b-a}{n}$$

In this case, $$f(x) = \dfrac{1}{x^2}$$

$$= f'(x) = -2x^{-3}$$

$$\Rightarrow f''(x) = 6x^{-4} = \dfrac{6}{x^4}$$

a=1 , b=4

$$\rightarrow$$ maximum value of $$f''(x)$$ in [1,4] is 6.

$$\Rightarrow k = 6$$

n=8,

$$\therefore |E_T| \leq \dfrac{6× (4-1)^3}{12× 64} \\= \dfrac{6× 3^3}{12× 64}\\ = 0.2109$$

$$\therefore |E_T| \leq 0.2109$$

### Estimate the error ,while calculating T8 for the integral $$I = \int\limits ^4_1 \dfrac{1}{x^2} dx$$.

A

$$|E_T| \leq 0.0156$$

.

B

$$|E_T| \leq 0.2109$$

C

$$|E_T| \leq 0.258$$

D

$$|E_T| \leq 5.16$$

Option B is Correct

# Error Bounds For Simpsons Rule

• Let ES be the error while calculating the value of integral using  Simpson Rule.
• Simpson Rule says that,

Simpsons rule for approximating integrals make use of parabolas instead of straight line for approximating curves we divide the intervals [a,b] into n equal parts,

each part $$=\Delta x=\dfrac{b-a}{n}$$ where n is even, then for each pair of intervals, we approximate y = f(x) >0 by a parabola as shown in the figures.    We take the interval $$[x_0, x_2]$$ where, $$x_0=-h, x_1=0, x_2=h$$the equation of parabola through P0, P1 & P2 is of the form y = Ax2+Bx +C  (where A,B,C are constants)

The area = $$\int\limits_{-h}^h(Ax^2 + Bx + C)dx=\dfrac {h} {3}(2Ah^2 + 6C)$$      .............(1)

Now the parabola passes through $$P_0(-h,y_0), P_1(0,y_1), P_2 (h,y_2)$$

$$\therefore$$$$y_0=Ah^2 - Bh + C\\y_1=C\\y_2=Ah^2 + Bh + C$$           ($$\therefore y_0 + 4y_1 + y_2=2Ah^2+6C$$)

put this in 1 we get,

Area of parabola = $$\dfrac{h}{3}(y_0+4y_1+y_2)$$

Similarly, area under the parabola $$P_2, P_3,P_4$$ will be $$\dfrac{h}{3}(y_2 + 4{y_3} + y_4)$$

$$\therefore$$Sum of all areas = $$\dfrac{h}{3}\Bigg[y_0+4y_1+y_0\Bigg] + \dfrac{h}{3}\Bigg[y_2+4y_3+y_4\Bigg] + \dfrac{h}{3}\Bigg[y_4+4y_5+y_0\Bigg] + .....+ \dfrac{h}{3}\Bigg[y_{n-2}+4y_{n-1}+y_n\Bigg]$$$$\dfrac{h}{3}\Bigg[y_0 + 4y_1 + 2y_2 + 4y_3 + 2y_4 + .............+2y_n-2 + 4y_{n-1} + y_n\Bigg]$$

$$\simeq\int\limits_a^bf(x)dx$$

$$\therefore \,\int\limits_a^bf(x)dx =S_n=\dfrac{\Delta x}{3}\Bigg[f(x_0) + 4f(x_1) + 2f(x_2) + ........+ 2f(x_{n-2}) + 4f(x_{n-1}) + f(x_n)\Bigg]$$

where, n is even and $$\Delta x=\dfrac{b-a}{n}$$

• Let $$|f'' (x)| \leq k$$, ( $$f''''(x)$$ is the fourth derivation  of $$f$$ w.r.t. $$x$$)

for $$a\leq x\leq b$$ then,

$$|E_S| \leq \dfrac{k(b-a)^5}{180\,n^4}$$

• The R.H.S. value in this inequality is the maximum error that can creep in.

• In general the actual error is substantially less than maximum error.

#### Estimate the  error while calculating  S8 for the integral  $$I = \int\limits^2_1 \dfrac{1}{x^2} dx$$

A $$|E_S| \leq 0.000012$$

B $$|E_S| \leq 0.000163$$

C $$|E_S| \leq 0.23$$

D $$|E_S| \leq 1.1136$$

×

For the Simpsons Rule

$$|E_S| = Error \leq \dfrac{k(b-a)^5}{180\, n^4}$$

where, $$|f''''(x)| \leq k\,\, \text{for}\,\,a\leq x\leq b$$

For the integral,  $$I = \int\limits^b_a f(x) dx$$ and  $$\Delta x= \dfrac{b-a}{n}$$

In this case,

$$f(x) = \dfrac{1}{x^2}$$ , $$\Rightarrow f'(x) = -2x^{-3}$$

$$\Rightarrow f''(x) = 6\,x^{-4}$$

$$\Rightarrow f'''(x) = -24\,x^{-5}$$

$$\Rightarrow f'''' (x) = 120\, x^{-6}$$

a=1 , b=2

The maximum value of  $$f^4(x) = 120$$.

$$\Rightarrow k=120$$

$$n=8$$,

$$\therefore |E_S| \leq \dfrac{k(b-a)^5}{180\,n^4} \\= \dfrac{120× (2-1)^5}{180× 8^4} \\= 0.000163$$

$$= |E_S| \leq 0.000163$$

### Estimate the  error while calculating  S8 for the integral  $$I = \int\limits^2_1 \dfrac{1}{x^2} dx$$

A

$$|E_S| \leq 0.000012$$

.

B

$$|E_S| \leq 0.000163$$

C

$$|E_S| \leq 0.23$$

D

$$|E_S| \leq 1.1136$$

Option B is Correct

# Approximation of Integral in which Function is given in Tabular Form :

• Sometimes to find the approximate value of $$I = \int\limits ^b_a f(x) dx$$ we are not given the formula for $$'f'$$ but a table which mention some values of $$x$$ and $$f(x)$$ for those corresponding values.
• We can use any of the three rules  Midpoint Rule, Trapezoidal  Rule  or Simpson Rule to estimate the value of this integral.
• We have to choose $$n$$  according to the data given .

#### Use Midpoint Rule and given data to estimate the value of  $$I = \int \limits^6_2 f(x) dx$$. $$x$$ $$f(x)$$ $$x$$ $$f(x)$$ 2.0 5.6 4.5 10.6 2.5 6.3 5.0 11.2 3.0 7.2 5.5 12.3 3.5 8.1 6.0 13.5 4.0 8.9

A 102.58

B 1.63

C 37.30

D 15.2

×

Midpoint rule says that,

$$\int\limits_a^bf(x)dx\cong\,\Delta\,x\,\Bigg[f(\overline x_1)+f(\overline x_2) + .........f(\overline x_n)\Bigg]$$

$$=M_{n}$$

where, $$\Delta\,x=\dfrac{b-a} {n}$$ and  $${\over{x_i}}=\dfrac{1}{2}(x_{i–1}+x_i)$$

In this problem,

we take n= 4 ,a = 2 , b= 6,

$$\Delta x = \dfrac{6-2}{4} = 1$$ , $$\bar x_1 = \dfrac{2+3}{2} = \dfrac{5}{2}$$$$\bar x_2 = \dfrac{3+4}{2} = \dfrac{7}{2}$$,$$\bar x_3 = \dfrac{4+5}{2} = \dfrac{9}{2}$$$$\bar x_4 = \dfrac{5+6}{2} = \dfrac{11}{2}$$

$$\therefore \int\limits^6_2 f(x) dx \simeq M_4 = \Delta x\Bigg[ f(\bar x_1)+f(\bar x_2)+f(\bar x_3)+f(\bar x_4)\Bigg]$$

Now, read the values of function from the table given

$$=1\Bigg[ f(2.5)+f(3.5)+f(4.5)+f(5.5)\Bigg]$$

$$= 1 \Bigg[6.3+8.1+10.6+12.3\Bigg]$$

$$= 37.3$$

### Use Midpoint Rule and given data to estimate the value of  $$I = \int \limits^6_2 f(x) dx$$. $$x$$ $$f(x)$$ $$x$$ $$f(x)$$ 2.0 5.6 4.5 10.6 2.5 6.3 5.0 11.2 3.0 7.2 5.5 12.3 3.5 8.1 6.0 13.5 4.0 8.9

A

102.58

.

B

1.63

C

37.30

D

15.2

Option C is Correct

# Approximation of Integral in Which Function is Given in Tabular Form

• Sometimes to find the approximate value of $$I = \int\limits ^b_a f(x) dx$$ we are given the formula for $$'f'$$ but a table which mentation same values of $$x$$ and $$f(x)$$ for those corresponding values.
• We can use any of the three rules Midpoint Rule, Trapezoidal  Rule  or Simpson Rule to estimate the value of this integral.
• We have to choose $$n$$  according to the data given .

#### Use Trapezoidal Rule and give data to estimate the value of $$I= \int\limits^6_2 f(x) dx$$ $$x$$ $$f(x)$$ $$x$$ $$f(x)$$ 2.0 5.6 4.5 10.6 2.5 6.3 5.0 11.2 3.0 7.2 5.5 12.3 3.5 8.1 6.0 13.5 4.0 8.9

A 15.82

B 36.85

C 1.93

D 105.62

×

By trapezoidal rule of approximation,

$$\int\limits_a^b\,f(x) \simeq \dfrac{\Delta x}{2} \Bigg[f(x_0) + 2f(x_1) + 2f(x_2) + .......+ 2f(x_{n–1}) + f(x_n)\Bigg]$$

$$=T_n$$

Where,  $$\Delta x = \dfrac {b-a}{n},x_i=a+i\Delta x$$

In this case,

we take $$n= 4$$ , a=2 , b=6

$$\Delta x= \dfrac{6-2}{4} = 1,\,x_0 = 2,\,x_1= 3, x_2= 4,x_3= 5,x_4=b= 6$$

$$\therefore \, \int\limits^6_2 f(x) dx \simeq T_4 = \dfrac{\Delta x}{2}\Bigg[f(x_0)+2f(x_1)+2f(x_2)+2f(x_3)+f(x_4)\Bigg]$$

$$= \dfrac{1}{2} \Bigg[f(2)+2f(3)+2f(4)+2f(5)+f(6)\Bigg]$$

$$= \dfrac{1}{2} \Bigg[5.6+2× 7.2+2× 8.9+2× 11.2+13.5\Bigg]$$

$$= \dfrac{1}{2} \Bigg[73.7\Bigg] = 36.85$$

### Use Trapezoidal Rule and give data to estimate the value of $$I= \int\limits^6_2 f(x) dx$$ $$x$$ $$f(x)$$ $$x$$ $$f(x)$$ 2.0 5.6 4.5 10.6 2.5 6.3 5.0 11.2 3.0 7.2 5.5 12.3 3.5 8.1 6.0 13.5 4.0 8.9

A

15.82

.

B

36.85

C

1.93

D

105.62

Option B is Correct

# Approximation of Integral in which Function is given in Tabular Form

• Sometimes to find the approximate value of $$I = \int\limits ^b_a f(x) dx$$ we are given the formula for $$'f'$$ but a table which mentation same values of $$x$$ and $$f(x)$$ for those corresponding values.
• We can use any of the three rules Midpoint Rule, Trapezoidal  Rule  or Simpson Rule to estimate the value of this integral.
• We have to choose $$x$$  according to the data given .

#### Use Simpson Rule and give data to estimate the value of $$I= \int\limits^6_2 f(x) dx$$ $$x$$ $$f(x)$$ $$x$$ $$f(x)$$ 2.0 5.6 4.5 10.6 2.5 6.3 5.0 11.2 3.0 7.2 5.5 12.3 3.5 8.1 6.0 13.5 4.0 8.9

A 59.62

B 36.83

C 0.02

D 511.23

×

Simpsons Rule says that ,

$$\int\limits_a^bf(x)dx \simeq S_n=\dfrac{\Delta x}{3} \Bigg[f(x_0) + 4f(x_1) + 2f(x_2) + .......2f(x_{n-2})+4f(x_{n-1}) + f(x_n) \Bigg]$$

Where, $$\Delta x=\dfrac{b-a}{n}$$ and n is even, $$x_i=a+i\Delta x$$

In this case,

we take n= 4 , a=2 , b=6,

$$\Delta x= \dfrac{6-2}{4}=1,\, x_0= 2, \,x_1= 3\,,\,x_2= 4\,,\,x_3= 5\,,\,x_4= b=6$$

$$\therefore \, \int\limits^6_2 f(x) dx \simeq S_4 = \dfrac{\Delta x}{3}\Bigg[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+f(x_4)\Bigg]$$

$$= \dfrac{1}{3} \Bigg[f(2)+4\,f(3)+2\,f(4)+4\,f(5)+f(6)\Bigg]$$

$$= \dfrac{1}{3} \Bigg[5.6+4× 7.2+2× 8.9+4× 11.2+13.5\Bigg]$$

$$= \dfrac{1}{3} \Bigg[5.6+ 28.8+17.8+44.8+13.5\Bigg]$$

$$= \dfrac{1}{3} \Bigg[110.5\Bigg]\\ = 36.83$$

### Use Simpson Rule and give data to estimate the value of $$I= \int\limits^6_2 f(x) dx$$ $$x$$ $$f(x)$$ $$x$$ $$f(x)$$ 2.0 5.6 4.5 10.6 2.5 6.3 5.0 11.2 3.0 7.2 5.5 12.3 3.5 8.1 6.0 13.5 4.0 8.9

A

59.62

.

B

36.83

C

0.02

D

511.23

Option B is Correct